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love history [14]
3 years ago
13

Sarah's water bottle holds 500 milliliters. Shawn container holds 4/5 as much water. How many total milliliters of water do both

containers hold put together
Mathematics
1 answer:
tamaranim1 [39]3 years ago
6 0
900 milliliters combined
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Will mark BRAINLIEST!
beks73 [17]

Hello There!

The first one seems alright.

The second one it would be 1/2.

This is because half of the options have “R” on them.

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3 years ago
Find a unit vector that is perpendicular to the vector 5, −4
konstantin123 [22]

( \frac{4}{ \sqrt{41} } ,  \frac{5}{ \sqrt{41} } )
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3 years ago
Eduardo worked 40 regular hours last week plus 7 hours over time at time and a half his gross pay was 757.50 what is his hourly
Diano4ka-milaya [45]

Answer:

divide

Step-by-step explanation:

and youll find the answer

4 0
2 years ago
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Let f(x) = [infinity] xn n2 n = 1. find the intervals of convergence for f. (enter your answers using interval notation. ) find
inna [77]

Best guess for the function is

\displaystyle f(x) = \sum_{n=1}^\infty \frac{x^n}{n^2}

By the ratio test, the series converges for

\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{(n+1)^2} \cdot \frac{n^2}{x^n}\right| = |x| \lim_{n\to\infty} \frac{n^2}{(n+1)^2} = |x| < 1

When x=1, f(x) is a convergent p-series.

When x=-1, f(x) is a convergent alternating series.

So, the interval of convergence for f(x) is the <em>closed</em> interval \boxed{-1 \le x \le 1}.

The derivative of f is the series

\displaystyle f'(x) = \sum_{n=1}^\infty \frac{nx^{n-1}}{n^2} = \frac1x \sum_{n=1}^\infty \frac{x^n}n

which also converges for |x| by the ratio test:

\displaystyle \lim_{n\to\infty} \left|\frac{x^{n+1}}{n+1} \cdot \frac n{x^n}\right| = |x| \lim_{n\to\infty} \frac{n}{n+1} = |x| < 1

When x=1, f'(x) becomes the divergent harmonic series.

When x=-1, f'(x) is a convergent alternating series.

The interval of convergence for f'(x) is then the <em>closed-open</em> interval \boxed{-1 \le x < 1}.

Differentiating f once more gives the series

\displaystyle f''(x) = \sum_{n=1}^\infty \frac{n(n-1)x^{n-2}}{n^2} = \frac1{x^2} \sum_{n=1}^\infty \frac{(n-1)x^n}{n} = \frac1{x^2} \left(\sum_{n=1}^\infty x^n - \sum_{n=1}^\infty \frac{x^n}n\right)

The first series is geometric and converges for |x|, endpoints not included.

The second series is f'(x), which we know converges for -1\le x.

Putting these intervals together, we see that f''(x) converges only on the <em>open</em> interval \boxed{-1 < x < 1}.

6 0
2 years ago
Art walked 4.1 km in preparation for the Community Walk–a–thon. Bill walked 3600 m. How many more metres did Art walk than Bill?
WINSTONCH [101]

Answer:

.5 more kilometres.

Step-by-step explanation:

3600 m = 3.6 km. 4.1 is .5 greater than 3.6

7 0
3 years ago
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